Answer
The solution of the equation $4{{x}^{2}}+8x+13=0$ in standard form is $\left\{ -1+\frac{3}{2}i,-1-\frac{3}{2}i \right\}$.
Work Step by Step
Consider the equation,$4{{x}^{2}}+8x+13=0$
Compare the equation $4{{x}^{2}}+8x+13=0$ with $a{{x}^{2}}+bx+c$.
$\begin{align}
& a=4 \\
& b=8 \\
& c=13
\end{align}$
Substitute $a=4$, $b=8$ and $c=13$ in the formula $x=\frac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
$\begin{align}
& x=\frac{-8\pm \sqrt{{{8}^{2}}-4\left( 4 \right)\left( 13 \right)}}{2\left( 4 \right)} \\
& =\frac{-8\pm \sqrt{64-208}}{8} \\
& =\frac{-8\pm \sqrt{-144}}{8}
\end{align}$
Use the property $\sqrt{-b}=i\sqrt{b}$.
\[\begin{align}
& x=\frac{-8\pm i\sqrt{144}}{8} \\
& =\frac{-8\pm 12i}{8} \\
& =\frac{-8}{8}\pm \frac{12}{8}i \\
& =-1\pm \frac{3}{2}i
\end{align}\]
Therefore, the solution of the equation $4{{x}^{2}}+8x+13=0$ in standard form is $\left\{ -1+\frac{3}{2}i,-1-\frac{3}{2}i \right\}$.
